6. Duals of L p spaces

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1 6 Duals of L p spaces This section deals with the problem if identifying the duals of L p spaces, p [1, ) There are essentially two cases of this problem: (i) p = 1; (ii) 1 < p < The major difference between these two cases is the fact that for 1 < p <, there is a nice characterization of the dual of L p which identifies it with L q (with 1 p + 1 q = 1), and this identification holds without any restriction on the underlying space On the other hand, the dual of L 1 will be identified with L loc, but this identification will work only if the underlying measure space is decomposable (eg σ-finite) Of course, one can also pose the problem of identifying the duals of the spaces L and L loc, introduced in Section 5, but we shall not deal with it We start off with the study of the finite case Comment The finite case is particularly nice, since among other things the L p spaces contain a blueprint of the underlying measure space More precisely, if (,, µ) is a finite measure space, then the characteristic functions κ,, belong to all spaces L p (,, µ), 1 p The following result somehow explains how the union can be seen in L p Lemma 61 Let (,, µ) be a finite measure space, let be either R or C, and let p [1, ) If ( n ) n=1 is a pairwise disjoint sequence, and if we put = n=1 n, then one has the equality n=1 κ n = κ in L p (,, µ) In other words, one has the equality κ = L p - lim f n, n where (f n ) n=1 L p (,, µ) is the equence of partial sums f n = n k=1 κ k, n 1 Proof One has f n = κ n, where n = 1 n, n 1 In particular, we get f n κ p dµ = κ n p dµ = κ n dµ = µ( n ), n 1 This computation can be re-written as (1) f n κ p = µ( n ) 1/p, n 1 Notice that using the inclusions 1 2, combined with the equality n=1 [ n] =, by the continuity property (it is key here that µ is finite), we get lim n µ( n ) = µ( ) = 0, so the equalities (1) immediately give lim n f n κ p = 0 n important consequence of this fact, we get our first duality result Theorem 61 Let (,, µ) be a finite measure space, let be either R or C, let p [1, ), and let φ : L p (,, µ) be a linear continuous map Define the number q = p/(p 1) (with the convention 1/0 = ) Then there exists some function f L q (,, µ), such that (2) φ(g) = fg dµ, g L p (,, µ) Moreover one has the following 300

2 6 Duals of L p spaces 301 (i) The function f is essentially unique, in the sense that, whenever f 1, f 2 L q (,, µ) satisfy (2), it follows that f 1 = f 2, µ-ae (ii) If f L q (,, µ) is any function satisfying (2), then f q = φ Proof We begin with the existence part Start off by considering the correspondence ν φ : φ(κ ) y Lemma 61, ν φ defines a -valued measure on Note also that ν φ is absolutely continuous with respect to µ Indeed, if has µ(), then κ = 0 in L p, so ν φ () = φ(κ ) = 0 Use the Radon-Nikodym Theorem (the finite case) to find some function f L 1 (,, µ), such that (3) φ(κ ) = fκ dµ, Since φ is linear, the equality (3) can be extended to give (4) φ(g) = fg dµ, g L p,elem (,, µ) We wish to extend the above equality beyond elementary functions Since (,, µ) is finite, one has an injective linear continuous map J p : L (,, µ) L p (,, µ), which comes from the inclusion L (,, µ) Lp (,, µ) Using the map J p, we identify L (,, µ) as a linear subspace in Lp (,, µ) (Strictly speaking, this subspace is Ran J p ) We also know that, one has the obvious inclusions L p elem, (,, µ) = L elem,(,, µ) L (,, µ) L p (,, µ), so in fact L (,, µ) is a dense linear subspace of Lp (,, µ) Claim 1: One has the equality (5) φ(g) = fg dµ, g L (,, µ) Consider the linear map S f : L (,, µ), defined by S f (g) = fg dµ, g L (,, µ) We know that S f is continuous (Proposition 55) Notice that the restriction ψ = φ gives rise to a linear continuous map ψ : L (,,µ) L (,, µ) (Strictly speaking, we have ψ = φ J p ) Then the equality (4) gives ψ(g) = fg dµ = S f (g), g L elem,(,, µ), and since both ψ and S f are continuous, and L elem, (,, µ) is dense in L (,, µ), we get the equality ψ = S f, which is precisely (5) Claim 2: The function f belongs to L q (,, µ)

3 302 CHPTER IV: INTEGRTION THEORY We start with the easy case p = 1 In this case we observe that, since κ 1 = µ(),, by the continuity of φ, we get ν φ () = φ(κ ) φ κ 1 = φ µ(),, so we can apply the easy Radon-Nikodym Theorem 41, to get the fact that f φ, µ-ae We continue with the case p > 1 We are going to give an estimate (from above) for f q dµ y definition, we know that { f q dµ = sup h dµ : h -Elem R (), 0 h f } q Fix for the moment an -elementary function h : R, such that 0 h f q Consider the measurable function u : by { f(x) /f(x) if f(x) 0 u(x) = 0 if f(x) = 0 so that one has the equality uf = f Consider now the function g = uh 1/p Since h 1/p is elementary, and, u 1, it follows that g is measurable, and bounded, hence in L y Claim 1, we then have φ(g) = fg dµ = fuh 1 p dµ = f h 1 p dµ mong other things, this proves that φ(g) = φ(g) 0, as well as (use f h 1 q ) the inequality φ(g) h 1 1 q h p dµ = h dµ y the continuity of φ, this inequality forces (6) h dµ φ g p Let us estimate the norm g p Since g p = u p h h, we clearly have [ ] 1 g p h dµ so if we denote h dµ simply by α, then (6) reads α φ α 1 p, so we immediately get α φ q Having proven the inequality h dµ φ q, for all elementary functions h, with 0 h f q, it now follows that f q is integrable, and f q dµ φ q, so f indeed belongs to L q (,, µ) (and moreover, f q φ ) Having proven the existence part, the other properties (i) and (ii) are clear from the results in Sections 3 and 5 p,

4 6 Duals of L p spaces 303 Our next goal is to extend the above result beyond the finite case In preparation for the proof in the general case, we need to develop a technique for breaking a linear continuous map φ : L p (,, µ) into (small) pieces For this purpose we introduce the following Notations Let (,, µ) be a measure space, let be one of the fields R or C, and let p [1, ) For any set, we define the map P : L p (,, µ) f f L p (,, µ) It turns out that P is surjective, linear, continuous, and it has a right inverse defined as follows For f L p (,, µ), we define the function f : by f = { f(x) if x 0 if x It turns out that f belongs to L p (,, µ), and f p = f p, so the correspondence gives rise to a linear continuous map L p (,, µ) f fl p (,, µ) R : L p (,, µ) L p (,, µ), which satisfies P R = Id Remark also that the composition the other way can be described as the multiplication map by κ (see Section 5): R P = M p κ Definitions Let (,, µ) be a measure space, let be one of the fields R or C, and let p [1, ) Given any linear continuous map φ : L p (,, µ), we define, for every the linear continuous maps One has the obvious equalities Notice also that one has the equality φ = φ R : L p (,, µ), φ = φ Mκ p : L p (,, µ) φ = (φ ) P and φ = φ R φ (f) = φ (fκ ), f L p (,, µ) Remark that, since P = R 1, one also has φ = φ φ If we take =, we clearly have φ = φ = φ Note also, that given sets, with, then one has (φ ) = φ and ( φ ) = φ In particular one has φ = φ φ = φ Remarks 61 Use the notations above Suppose 1,, n are pairwise disjoint, and let = 1 n For any linear continuous map φ : L p (,, µ), one has the equality φ = φ φ n

5 304 CHPTER IV: INTEGRTION THEORY This is a consequence of the fact that the correspondence is linear, so in particular one has L,loc (,, µ) f M p f L( L p (x,, µ)) M p κ = M p κ M p κ n For any g L p (,, µ), one has the equality This is quite clear, since ( ) p gκ p = gκ p dµ = gκ p = [( gκ 1 p ) p + + ( gκ n p ) p ] 1 p g p κ dµ = n k=1 g p κ k dµ = The following result gives an important property of these operations n ( ) p gκ k p Lemma 62 Let (,, µ) be a measure space, let be either R or C, and let, be disjoint sets For any linear continuous map φ : L 1 (,, µ), one has φ = max { φ, φ } If p, q (1, ) are such that 1 p + 1 q = 1, then for any linear continuous map φ : L p (,, µ), one has φ = [ φ q + + φ q] 1 q Proof One useful observation is the fact that, for every p [1, ) and every linear continuous map φ : L p (,, µ), one has (7) φ (g) = φ (g) + φ (g) = φ (gκ ) + φ (gκ ), g L p (,, µ) Let us remark first that, since (φ ) = φ and (φ ) = φ, we clearly have φ max { φ, φ } To prove the other inequality φ max { φ, φ }, we must show that (8) (φ )(g) max { φ, φ } g 1, g L 1 (,, µ) Start with some g L 1 (,, µ) Using (7), we clearly have φ (g) φ (gκ ) + φ (gκ ) φ gκ 1 + φ gκ 1 (9) max { φ, φ } ( ) gκ 1 + gκ 1 Using Remark 61 we also have gκ 1 + gκ 1 = gκ 1 g 1, and then using (9) we immediately get (8) We first prove the inequality (10) φ [ φ q + φ q] 1 q We assume φ, φ > 0 (otherwise there is nothing to prove) Start with some ε with 0 < ε < min{ φ, φ }, and choose two functions g, h L p (,, µ) with g p, h p 1 such that φ (g) φ ε and φ (h) φ ε k=1

6 6 Duals of L p spaces 305 Replacing g with φ (g) φ (g) g, we can assume that φ (g) = φ (g) Similarly, we can assume φ (h) = φ (h) Fix for the moment an arbitrary pair (α, β) [0, ) 2, with α p + β p = 1, and define the function f = αgκ + βhκ y construction we have f = fκ, as well as fκ = αgκ and fκ = βhκ, so by Remark 61, we have so we get f p = [( αgκ p ) p + ( βhκ p ) p ] 1 p [( αg p ) p + ( βh p ) p ] 1 p [α p + β p ] 1 p = 1, φ φ (f) Notice that, since κ κ = 0, clearly have have φ (f) = φ (αgκ ) = α φ (g) and φ (f) = φ (βhκ ) = β φ (h), which yields φ (f) = α φ (g) + β φ (g) α ( φ ε) + β ( φ ε) Since f p 1, by the definition of the norm, the above estimate forces φ α ( φ ε) + β ( φ ε) In the above inequality α, β [0, ) are arbitrary, subject to α p + β p = 1 In particular, if we consider numbers [ ( φ ε) q ] 1 [ p ( φ ε) q ] 1 p α = ( φ ε) q + ( φ ε) q and β = ( φ ε) q + ( φ ε) q, we get φ ( φ ε) q p +1 + ( φ ε) q p +1 [ ( φ ε) q + ( φ ε) q ] 1 p = ( φ ε) q + ( φ ε) q [ ( φ ε) q + ( φ ε) q ] 1 p = [ ( φ ε) q + ( φ ε) q ] 1 1 p = [ ( φ ε) q + ( φ ε) q ] 1 q Since the inequality φ [ ( φ ε) q + ( φ ε) q ] 1 q holds for every ε (0, min{ φ, φ }), it will clearly force (10) To prove the other inequality φ [ φ q + φ q] 1 q, we must show that (11) (φ )(g) [ φ q + φ q] 1q g p, g L p (,, µ) Start with some g L p (,, µ) Using (7), combined with Hölder s inequality, we have φ (g) φ (gκ ) + φ (gκ ) φ gκ p + φ gκ p (12) [ φ q + φ q] [ 1q ( ) p ( ) ] 1 p p gκ p + gκ p Using Remark 61 we also have [ ( gκ p ) p + ( gκ p ) p ] 1 p = gκ p g p, =

7 306 CHPTER IV: INTEGRTION THEORY and then using (12) we immediately get (11) Theorem 62 (Finite approximation) Let (,, µ) be a measure space, let be either R or C, and let p [1, ) For every linear continuous map φ : L p (,, µ), one has (13) φ = sup { φ :, µ() < } Moreover, if p > 1, one also has (14) inf { φ φ :, µ() < } = 0 Proof It is obvious that φ sup { φ :, µ() < } To prove the other inequality, we start with some ε > 0 Since L p elem, (,, µ) is dense in L p (,, µ), there exists f Lp elem, (,, µ), with f p 1, and φ(f) φ ε Notice that the fact that f L p elem, (,, µ) means that f = λ 1 κ λ n κ n, with k and µ( k ) <, k = 1,, n In particular, if take = 1 n then, µ() <, and we also have f = fκ, which gives φ (f) = φ(f) Therefore, we get φ ε φ (f) φ f p = φ, thus proving that we have in fact equality in (13) To prove the equality (??), we use (13) to find, for every integer n 1, some set n, with µ( n ) <, such that (15) φ n > φ 1 n Since we clearly have φ n = φ φ n, by Lemma 62 we get where q = p/(p 1) This yields and using (15) we get φ = [ φ n q + φ φ n q] 1 q, φ φ n = [ φ q φ n q] 1 q, φ φ n [ φ q ( φ 1 n which clearly forces lim n φ φ n = 0 ) q ] 1 q, Remark 62 The equality (??) does not holds for p = 1, the reason being the fact that in this case one only has φ φ = φ For example, suppose one works with a σ-finite measure space (,, µ), with µ() =, and if one considers the integration map φ : L 1 (,, µ) f f dµ, then it is pretty clear that φ = 1,, with µ() > 0 In particular, we see that for this map one has inf { φ φ :, µ() < } = inf { φ :, µ() < } = 1

8 6 Duals of L p spaces 307 Exercise 1 Suppose (,, µ) is a measure space, and p [1, ) Prove that, for every linear continuous map φ : L p (,, µ), there exists a µ-σ-finite set, with φ = φ Moreover, if p > 1, then any set with this property forces the equality φ = φ We are now ready to prove our first duality result efore we do so, let us recall some notations and results from Section 3 Notations Let (,, µ) be a measure space, let be one of the fields R or C, and let p, q (1, ) be Hölder conjugate numbers, ie 1 p + 1 q = 1 For any f L q (,, µ) we consider the map Λ f : L p (,, µ) g fg dµ We know from the results in Section 3 that for every f L q (,, µ), the map Λ f continuous, and has Λ f = f q the correspondence (16) Λ : L q (,, µ) f Λ f L p (,, µ) is linear continuous and isometric : L p (,, µ) is linear Theorem 63 (Duality Theorem for L p, p (1, )) With the notations above, the correspondence (16) is an isometric linear isomoprhism Proof ll we need to prove is the surjectivity of (16) Start with some linear continuous map φ : L p (,, µ), and let us construct a function f L q (,, µ), such that Λ f = φ The key step is contained in the following Claim 1: For every with µ() <, there exists some f L q (,, µ), such that Λ f = φ Indeed, if one considers the linear continuous map φ : L p (,, µ), then by Theorem 61, there exists a function f 0 L q (,, µ) with (φ )(g) = f 0 g dµ, g L p (,, µ) If we define the function f = R f 0 L q (,, µ), then we have f = f 0 and f = fκ, so for every g L p (,, µ) we have fg dµ = f 0 (g ) dµ = (φ )(g ) = φ (g) Having proven Claim 1, we now use Theorem 62 to find a sequence ( n ) n=1, with µ( n ) <, n 1, such that lim n φ φ n = 0 For each n 1 we use Claim 1 to find some f n L q (,, µ) such that Λ f n = φ n, so now we have (17) lim n φ Λ f n = 0 Claim 2: The sequence (f n ) n=1 L q (,, µ) is Cauchy Indeed, from (17) it follows that the sequence (Λ fn ) n=1 is Cauchy in the anach space L p (,, µ), ie lim N [ sup { Λfm Λf n : m, n N }] = 0

9 308 CHPTER IV: INTEGRTION THEORY Since Λ is linear and isometric, we have Λ fm Λ fn = Λ fm f n = f m f n q, m, n 1, so the above estimate gives [ { sup fm f n q : m, n N }] = 0 lim N Use now the fact that L q (,, µ) is a anch space, to conclude that L q - lim n f n = f, for some f L q (,, µ) Since Λ is isometric, we have and then (17) forces φ = Λ f lim Λ f Λ fn = 0, n efore we deal with the duality problem for L 1, we recall some notations and results from Section 5 Notations Let (,, µ) be a measure space, let be one of the fields R or C For any f L,loc (,, µ) we consider the map Λ f : L 1 (,, µ) g fg dµ We know from the results in Section 5 that for every f L,loc (,, µ), the map Λ f : L 1 (,, µ) is linear continuous, and has Λ f = f loc the correspondence (18) Λ : L,loc (,, µ) f Λ f L 1 (,, µ) is linear continuous and isometric Theorem 64 (Duality Theorem for L 1 ) Use the notations above, and assume there exists 0 such that (i) the measure space ( 0, 0, µ) is decomposable; (ii) the measure space ( 0, 0, µ) is degenerate, ie µ() {0, }, for all with 0 Then the correspondence (18) is an isometric linear isomoprhism Proof s before, all we have to show is the surjectivity of the map (??) First of all, since the measure space ( 0, 0, µ) is degenerate, it is clear that the restriction maps L 1 (,, µ) f f 0 L 1 ( 0, 0, µ) L,loc (,, µ) f f 0 L,loc ( 0, 0, µ) are isometric linear isomorphisms, so we can assume that in fact we have 0 = Recall that the decomposability condition means that there exists a collection F with µ(f ) <, F F; F F F = ; if a set has F, F F, then ; for every with µ() <, one has µ() = F F µ( F )

10 6 Duals of L p spaces 309 To prove the surjectivity of (??) we start with some linear continuous map φ : L 1 (,, µ), and we wish to find some function f L,loc (,, µ), with φ = Λ F The following first step is proven the exact same way as the first step in the proof of Theorem 62 Claim 1: For every with µ() <, there exists g L,loc (,, µ), such that Λ g = φ To construct the desired function f, we use the above Claim to find, for each F F a function f F L,loc (,, µ), such that Λ ff = φ F Using Lemma 51, we can in fact assume that f F is in fact bounded, with f F sup = f F loc Remark that, using the fact that (18) is isometric, it follows that, for every F F, one has the inequality (19) f F sup = Λ ff = φ F φ Use now the patching property (see??) to produce a measurable function f : such that f F = f F, F F Obviously f is bounded, so in particular it belongs to L,loc (,, µ), and one has the equalities (20) (Λ f ) F = Λ ff = φ F, F F Claim 2: For any set with µ() <, one has the equality Λ f (κ ) = φ(κ ) To prove this we first observe that, using decomosability, we have µ( F ) = µ() <, F F so in particular the collection F = {F : µ( F ) > 0 } is at most countable If we list it as a sequence F = {F 1, F 2, } (finite or infinite) then µ() = n µ( F n ), and it is pretty clear (use Remark 61 for example) that one has κ = L 1 - n κ Fn (The sum in the right hand side is either an ordinary sum, if F is finite, or a series, convergent in L 1, if F is infinite) In any event, using the equalities κ F = 0, F F F, the continuity of both Λ f and φ will give the fact that we have the equalities Λ f (κ ) = Λ f (κ F ) and φ(κ ) = φ(κ F ) F F F F This means that, in order to prove the Claim, all we have to show are the equalities Λ f (κ F ) = φ(κ F ), F F ut these equalities immediately follow from (20): Λ f (κ F ) = Λ f (κ κ F ) = (Λ f ) F (κ ) = φ F (κ ) = φ(κ F ) We now prove the equality Λ f = φ Using Claim 2, and linearity, it follows that (21) Λ f (g) = φ(g), g Span ( {κ :, µ() < } ) Since Span ( {κ :, µ() < } ) = L 1 elem, (,, µ) is dense in L1 elem, (,, µ), the continuity of both Λ f and φ, combined with (21) will force Λ f = φ

11 310 CHPTER IV: INTEGRTION THEORY Comment The conclusion of Theorem 64 may not be true in general Heuristically this can be seen in the following situation (the example will follow) Suppose we have a decomposable measure space (, M, µ) (on which Theorem 64 is true!), and we pick a small σ-algebra M If we consider the measure space (,, µ ), then by Exercise 6 in Section 4 we have an inclusion L 1 (,, µ ) L 1 (, M, µ), so if one starts with some f 0 L,loc (, M, µ), we get a linear continuous map φ = Λ M L f 0 1 (We use supercripts to indicate the corresponding measure space) It may happen that we have no way of (,,µ ) representing φ as Λ f with f L,loc (,, µ ), simply because the only candidate might be f = f 0 Example 61 Let be an uncountable set, and let µ : P() [0, ] be the counting measure For the measure space (, P(), µ) we have the obvious identification L 1 (, P(), µ) l1 (), with the integration corresponding to the map l 1 () g g(x) x Consider the σ-algebra = { : either or is countable } Remark that, although P(), one has the equality L 1 (,, µ ) = l 1 () Remark also that both measure spaces (, P(), µ) and (,, µ ) have the finite subset property, and in fact one has the equivalence (µ-lae) (µ-ae) (everywhere) Let be a set which does not belong to, and consider the map φ : L 1 (,, µ) g x g(x) We claim that there is no function f L (,, µ ) such that φ(g) = fg d(µ ), g L 1 (,, µ ) Indeed, if such a function exists, then if one considers the functions like κ {x}, x, which clearly belong to L 1 (,, µ ), then we immediately get { 1 if x f(x) = φ(κ {x} ) = 0 if x ie f = κ, which is impossible, since Exercise 2 Let (,, µ) be a measure space, and let be a σ-algebra Consider the measure space (, µ ) Let be either R or C s seen in Exercise 6 from Section 3, for every q [1, ), one has the inclusion L q (,, µ ) L q (,, µ) ssume 1 < q <, and let p = q/(q 1) Prove that for every f L q (,, µ), there exists a (unique µ-ae) function f L q (,, µ ) such that fg dµ = fg, dµ, g L p (,, µ )

12 6 Duals of L p spaces 311 Prove that, if the measure space (,, µ ) is decomposable, then for every f L 1 (,, µ), there exists a (unique µ-ae) function f L 1 (,, µ ) such that fg dµ = fg, dµ, g L,loc (,, µ ) Hints: For each f L q (,, µ) consider the map Φ f : L p (,, µ ) g fg dµ, and use Theorem 63 applied to the measure space (,, µ ) Use the Radon-Nikodym Theorem, on the measure space (,, µ ) for the -valued measure ν f () = fκ dµ, Comment Using the notations from the above Exercise, (if q = 1 we also require the extra hypothesis in part ) the map E q : Lq (,, µ) f f L q (,, µ ) is called the conditional expectation map This construction is often employed in Probability Theory In the case q = 2 the map E 2 can also be described as the orthogonal projection of L 2 (,, µ) onto the closed linear subspace L2 (,, µ ) Exercise 3 Use the notations from Exercise 2 (If q = 1 assume also the same hypothesis as in part ) Prove that the conditional expectation map E q : Lq (,, µ) Lq (,, µ ) is linear, continuous, and has E q 1 lso prove that (i) E q (f) = f, f Lq (,, µ ); (ii) E q (h f) = h Eq (f), h L,loc (,, µ ), f L q (,, µ) Conditional expectations can also be defined on L,loc Exercise 4 Use the notations from Exercise 2 Prove that, if the measure space (,, µ ) is decomposable, then for every f L,loc (,, µ), there exists a (unique µ-lae) function f L,loc (,, µ ) such that fg dµ = fg, dµ, g L 1 (,, µ ) Prove that the map E : L,loc (,, µ) f f L,loc (,, µ ) is again linear continuous and contractive Moreover, it also satifies the properties: (i) E (f) = f, f L,loc (,, µ ); (ii) E (h f) = h E (f), h L,loc (,, µ ), f L,loc (,, µ) Hint: For the existence of f argue as in the hint to Exercise 2, but use Theorem 64 instead

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